To determine the zeros of the quadratic function \( y = x^2 + 15x + 26 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = 15 \), and \( c = 26 \).
First, we need to calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 15^2 = 225 \]
\[ 4ac = 4 \times 1 \times 26 = 104 \]
Now, compute the discriminant:
\[ b^2 - 4ac = 225 - 104 = 121 \]
Since the discriminant is positive, we will have two distinct real zeros.
Now, substitute back into the quadratic formula:
\[ x = \frac{-15 \pm \sqrt{121}}{2 \cdot 1} \]
Simplifying this further:
\[ \sqrt{121} = 11 \]
Now we have:
\[ x = \frac{-15 \pm 11}{2} \]
Calculating the two values:
- For \( x = \frac{-15 + 11}{2} \):
\[ x = \frac{-4}{2} = -2 \]
- For \( x = \frac{-15 - 11}{2} \):
\[ x = \frac{-26}{2} = -13 \]
Thus, the two zeros of the quadratic function \( y = x^2 + 15x + 26 \) are:
\[ x = -13 \quad \text{and} \quad x = -2 \]
In summary, the smaller zero is \( -13 \) and the larger zero is \( -2 \).
Therefore, the final answer is:
\[ \text{Zeros: } -13, -2 \]